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Behind a square wall of side 10 m, two non-mixing liquids of densities rho1 and rho2 (with rho2 = 2*rho1) are held, the lighter on top. Each liquid forms a layer 5 m deep. Let M, N and O be points on the wall from top to bottom (M at top, N at the 5 m interface, O at the bottom). Find the ratio of the thrust on the upper half MN to the thrust on the lower half NO of the wall.
- 1/4
- 2/3
- 1/3
- 1/2
Correct answer: 1/4
Solution
Upper part MN (depth 0 to 5 m, density rho1): force F1 = rho1*g*(width)*integral of depth = rho1*g*W*(5²/2). Lower part NO (depth 5 to 10 m): pressure = rho1*g*5 (from top layer) + rho2*g*(d-5) for d in [5,10]. With rho2 = 2*rho1 and width W = 10, integrating gives F2. The ratio F1/F2 works out to 1/4.
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